Chapter 15 - Electric charges and fields

15.1Overview¶

In this and subsequent chapters, we start to look at the theories that describe electric and magnetic phenomena. Within the framework for dynamics that was developed by Newton, we will introduce the theories of electromagnetism which describe the electric force, the magnetic force, and how these two interact. This first chapter introduces the description of the electric force, analogously to how we introduced Newton’s Universal Theory of Gravity to describe the gravitational force.

15.2Electric charge¶

You have likely experienced or heard about electric charge in your life. For example, on a dry winter day, you might find that after rubbing your bare feet on a polyester carpet, you feel a small electric shock upon touching a metallic surface such as a doorknob. You probably also know that there are positive and negative charges, and that equal charges repel each other whereas opposite charges attract. In this chapter, we develop the description of how these charges can accumulate and how they exert attractive or repulsive forces on each other.

Ordinary matter is made of atoms, which are themselves made of a small positive nucleus (containing positive protons and neutral neutrons) surrounded by a “cloud” of negatively charged electrons. Within a solid object, inter-atomic forces hold the atoms together, so we can model the atoms as being effectively stationary. This means that we can treat the positive nuclei as being fixed in space. The negative electrons, depending on the material, can often move from one atom to another. If an atom loses an electron to another atom, it becomes positive, whereas the atom that acquired the extra electron becomes negative.

We define the net charge on an atom (or an object) based on whether there are more protons (positive), more electrons (negative) or an equal amount (neutral). By default, atoms are neutral and have an equal number of protons and electrons. An object becomes charged when it acquires an excess (or deficit) of electrons from another object. We say that “charge is conserved” because the number of electrons in the system does not change, i.e. if one object became positively charged, a different object must have become negatively charged by the same amount, so that the total net charge (in the Universe) is zero.

When you rub two objects together, you allow electrons to be transferred between them. Different materials have different “affinities” for electrons, and the electrons will transfer to the object with the highest affinity. Returning to our earlier example, when you rub your bare feet on the polyester carpet, electrons are being removed from your feet and deposited onto the carpet. Since your feet lose electrons, you acquire a net positive charge, while the carpet acquires a net negative charge. This way of creating a net charge on an object is called “charging by friction”. If you then touch a doorknob, electrons will jump from the doorknob and onto your (now positively charged) body, creating a spark.

The “triboelectic series” is a list of materials that tend to give up or acquire electrons when they are placed in close contact with each other. Some common materials from the series are shown in Figure 1. The greatest charge is generated by rubbing together materials that are the furthest apart from each other in the diagram. Rubbing silk on a piece of glass results in more charge than rubbing wool on the same piece of glass.

15.2.1Conductors and insulators¶

We can broadly classify materials into conductors (such as metals), and insulators (such as wood), depending on how easily the electrons can move around in the material. In a conductor, electrons (rather, the outer electron(s) of an atom) are only loosely bound to their nuclei, and they can thus move around the material freely. In an insulator, the electrons are tightly bound to the nuclei of their atoms and cannot easily move around. There is a third class of materials, semi-conductors, that falls somewhere between a conductor and an insulator. In a semi-conductor, electrons are typically bound to their atoms, but any additional electrons present in the material can move around as if they are in a conductor.

Within a conductor, such as a solid metallic sphere, charges can move around freely. If that sphere has a net charge, for example an excess of electrons, those excess electrons will migrate to the outer surface of the sphere. Electrons in the sphere repel each other and will quickly settle into a position where they are, on average, the furthest from all of the other electrons, which occurs if all of the electrons migrate to the surface. This is illustrated in the left panel of Figure 2. If an initially neutral conducting sphere is connected to the charged sphere by a conducting wire (right panel of Figure 2), some of the electrons will “conduct” (transfer) onto the surface of the neutral sphere, so that, on average, they are further from all other electrons. This way of adding charge to the neutral sphere is called “charging by conduction”, and the second sphere will remain charged if the connection between spheres is broken.

15.2.2Electrostatic induction¶

Electrostatic induction allows one to “induce” a charge by using the fact that charges can move freely in a conductor. The left panel of Figure 3 shows a (neutral) rod made of a conducting material, with electrons distributed uniformly throughout its volume. In the right panel, a negatively charged sphere is brought next to the rod. The negative sphere repels the electrons in the rod. Since the rod is conducting, the electrons can move around easily, and so they move to the end of the rod that is furthest from the negative sphere. Those electrons will leave positive charges (corresponding to the atoms that have lost their electrons) on the side closest to the sphere. The electrons in the rod will only accumulate for as long as the force from the negative sphere is greater than the repulsive force from the electrons that have already accumulated. In practice, such an equilibrium is reached almost instantly. In equilibrium, we say that the rod is “polarized”, or that the “charges in the rod have separated”, although the rod is overall still neutral.

Note that we can model this as if it where positive charges that move inside of the rod instead of negative charges. The positive charges are attracted to the negative sphere, and thus accumulate on the end of the rod closest to the sphere, leaving a negative charge on the other end. The choice to call electrons “negative” is completely arbitrary. For convenience, we usually build models assuming that positive charges can easily move around, even if, in reality, it is almost always actually (negative) electrons that move.

We can create a net charge on the polarized rod if we provide a conducting path for charges to leave (or enter) the rod. The Earth can be modelled as a very large reservoir of both positive and negative charges. By connecting the rod to the Earth (we say that we connect the rod to “ground”), we provide a path for the electrons in the rod to be even further from the negatively charged sphere, and they can thus leave the rod entirely in order to go into the ground. This is illustrated in the right-hand panel of Figure 4.

If we then disconnect the rod from the ground, it has now acquired an overall positive charge, as in the right hand panel. We call this “charging by induction”. We can also think of this in terms of positive charges moving into the rod from the Earth; when we connect the rod to the ground, the positive charges in the Earth can move into the rod and get closer to the negatively charged sphere. If we disconnect the rod from the ground, the rod stays positive, just as we conclude when using a model where it is the negative charges that move^[40].

15.3The Coulomb force¶

Coulomb was the first to provide a detailed quantitative description of the force between charged objects. Nowadays, we use the (derived) SI unit of “Coulomb” (C) to represent charge. The “charge” of an object corresponds to the net excess (or lack) of electrons on the object. An electron has a charge of $-e=-1.6\times 10^{-19} {\rm C}$. Thus, an object with a charge of $-1 {\rm C}$ has an excess of about $\frac{1}{1.6\times 10^{-19}}=6.25\times 10^{18}$ electrons on it, which is a very large charge. If an object has an excess of electrons, it is negatively charged and we indicate this with a negative sign on the charge of the object. An object with a (positive) charge of $1 {\rm C}$ thus has a deficit of $6.25\times 10^{18}$ electrons.

Through careful studies of the force between two charged spheres, Coulomb observed[^41] that:

• The force is attractive if the objects have opposite charges and repulsive if the objects have the same charge.
• The force is inversely proportional to the squared distance between spheres.
• The force is larger if the charges involved are larger.

This leads to Coulomb’s Law for the electric force (or simply “Coulomb’s Law”), $\vec F_{12}$, exerted on a point charge $Q_1$ by another point charge $Q_2$:

where $\hat r_{21}$ is the unit vector from $Q_2$ to $Q_1$ and $r$ is the distance between the two charges, as illustrated in Figure 5. Coulomb’s constant, $k=8.99\times 10^9 {\rm N\cdot m^2/C^{2}}$, is simply a proportionality constant to ensure that the quantity on the right will have units of Newtons when all other quantities are in S.I. units. In some instances, it is more convenient to use the “permittivity of free space”, $\epsilon_0$, rather than Coulomb’s constant, in which case Coulomb’s Law has the form:

where $\epsilon_0=\frac{1}{4\pi k}=8.85\times 10^{-12} {\rm C^2\cdot N^{-1}\cdot m^{-2}}$ is a more fundamental constant, as we will see in later chapters.

If the two charges have positions $\vec r_1$ and $\vec r_2$, respectively, then the vector $\hat r_{21}$ is given by:

Coulomb’s Law is mathematically identical to the gravitational force in Newton’s Universal Theory of Gravity. Rather than quantity of mass determining the strength of the gravitational force, it is the quantity of charge that determines the strength of the electric force. The only major difference is that gravity is always attractive, whereas the Coulomb force can be repulsive.

The product $Q_1Q_2$ in the numerator of Coulomb’s force is positive if the two charges have the same sign (both positive or both negative) and negative if the charges have opposite signs. Again, referring to Figure 5, if the two charges are positive, the force on $Q_1$ will point in the same direction as $\hat r_{21}$ (since all of the scalars are positive in Coulomb’s Law) and thus be repulsive. If, instead, the two charges have opposite signs, the product $Q_1Q_2$ will be negative and the force vector on $Q_1$ will point in the opposite direction from $\hat r_{21}$ and the force is attractive.

In the following trinket, a positive and negative charge are drawn. The force each charge experiences is drawn as an arrow and calculated for a distance of $1 \overset{\circ}{\rm A}=1\times 10^{-10} {\rm m}$ apart. Following convention, the vector arrows have their origin at the charge experiencing the force. When the trinket is run, the force arrows indicate an attractive force, i.e., the charges experience forces toward one another. If the charges are changed to like charges, e.g., Q1.q = q and Q2.q = q, the force arrows indicate repulsion between the charges. Try it in the trinket below. What happens to the size of the force vectors if you make one charge twice as big as the other? Can you support this result with one of Newton’s Laws?

15.4The electric field¶

We define the electric field vector, $\vec E$, in an analogous way as we defined the gravitational field vector, $\vec g$. By defining the gravitational field vector, say, at the surface of the Earth, we can easily calculate the gravitational force exerted by the Earth on any mass, $m$, without having to use Newton’s Universal Theory of Gravity. As you recall, we can define the gravitational field, $\vec g(\vec r)$, at some position, $\vec r$, from a point mass, $M$, as the gravitational force per unit mass:

where $\vec r$ is a vector from the position of $M$ to where we want to know the gravitational field. As a result, the force exerted on a “test mass”, $m$, located at position $\vec r$ relative to mass $M$ is given by:

which, of course, is the result from Newton’s Theory of Gravity. As you recall, we can define the gravitational field for any object that is not a point mass (e.g. the Earth), and use that field to find the force exerted by the Earth on any mass $m$, without having to re-calculate the gravitational field each time (which requires an integral or Gauss’ Law).

We proceed in an analogous way to define the “electric field”, $\vec E(\vec r)$, as the electric force per unit charge. If we have a point charge, $Q$, located at the origin of a coordinate system, then the electric field from that point charge, $\vec E(\vec r)$, at some position, $\vec r$, relative to the origin is given by:

If we place a “test charge”, $q$, at position $\vec r$ in space, it will experience a force given by:

just as we find from Coulomb’s Law.

In Example 15.2, we determined the electric force on charge $q$, exerted by two other charges $Q_1$ and $Q_2$. If we now changed the value of $q$ and wanted to determine the force, we can use the electric field to simplify the process considerably. That is, we can determine the value of the electric field, $\vec E$, from $Q_1$ and $Q_2$ at the position of $q$, and then simply multiply that field vector by a charge $q$ to obtain the force on that charge, without having to add force vectors.

15.4.1Visualizing the electric field¶

Generally, a “field” is something that has a different value at different positions in space. The pressure in a fluid under the presence of gravity is a field: the pressure is different at different heights in the fluid. Since pressure is a scalar quantity (a number), we call it a “scalar field”. The electric field is called a “vector field”, because it is a vector that is different at each position in space. One way to visualize the electric field is to draw arrows at different positions in space; the length of the arrow is then proportional to the strength of the electric field at that position, and the direction of the arrow represents the direction of the electric field. The electric field for a point charge is shown using this method in Figure 11.

One disadvantage of visualizing a vector field with arrows is that the arrows take up space, and it can be challenging to visualize how the field changes magnitude and direction continuously through space. For this reason, one usually uses “field lines” to visualize a vector field. Field lines are continuous lines with the following properties:

• The direction of the vector field at some point in space is tangent to the field line at that point.
• Field lines have a direction to indicate the direction of the field vector along the tangent (as there are two possibilities, parallel and anti-parallel).
• The magnitude of the field is proportional to the density of field lines at that point. The more field lines near a location in space, the larger the magnitude of the field vector at that point.

An example of using field lines to represent a vector field in space is shown in Figure 12. The corresponding field vector is shown at two different positions in space ($A$ and $B$). At both positions, the vector is tangent to the field line at that position in space and points in the direction of the little arrow drawn at the end of the field lines. The field vector at point $A$ has a larger magnitude than the one at point $B$, since the field lines are more concentrated at point $A$ than at point $B$ (there are more field lines per unit area at that position in space, the field lines are closer together).

Because the electric field vector always points in the direction of the force that would be exerted on a positive charge, electric field lines will point out from a positive charge and into a negative charge. The electric field lines for a combination of positive and negative charges is illustrated in Figure 13.

15.4.2Electric field from a charge distribution¶

So far, we have only considered Coulomb’s Law for point charges (charges that are infinitely small and can be considered to exist at a single point in space). We can use the principle of superposition to determine the electric field from a charged extended/continuous object by modelling that object as being made of many point charges. The electric field from that object is then the sum of the electric field from the point charges that make up that object.

Consider a charged wire that is bent into a semi-circle of radius $R$, as in Figure 14. The wire carries a net positive electric charge, $+Q$, that is uniformly distributed along the length of the wire. We wish to determine the electric field vector at the centre of the circle.

We start by choosing a very small section of wire and model that section of wire as a point charge with infinitesimal charge $dq$ (as in Figure 15). A distance $R$ from that point charge, the electric field from that point charge will have magnitude, $dE$, given by:

The electric field vector, $d\vec E$, from the point charge $dq$ is illustrated in Figure 15.

Using the coordinate system that is shown, we define θ as the angle made by the vector from the origin to the point charge $dq$ and the $x$-axis. The electric field vector from $dq$ is then given by:

The total electric field at the origin will be obtained by summing the electric fields from the different $dq$ over the entire semi-circle:

We are thus left with two integrals to solve for the $x$ and $y$ components of the electric field, respectively. Before jumping into solving the integrals, it is useful to think about the symmetry of the problem. Specifically, consider a second point charge, $dq'$, located symmetrically about the $x$-axis from charge $dq$, as illustrated in Figure 15. The charge $dq'$ will create a small electric field $d\vec E'$ as illustrated. When we add together $d\vec E$ and $d\vec E'$, the two $y$ components will cancel, and only the $x$ components will sum together. Similarly, for any $dq$ that we choose, there will always be another $dq'$ such that when we sum together their respective electric fields, the $y$ components will cancel. Thus, by symmetry, we can argue that the net $y$ component of the electric field, $E_y$, must be zero. We thus only need to evaluate the $x$ component of $\vec E$:

In order to solve this integral, we need to consider which variables change for different choices of the point charge $dq$. In this case, the distance $R$ is the same anywhere along the semi-circle, so only θ changes with different choices of $dq$, as $k$ is a constant. We can express $dq$ in terms of $d\theta$ and then use θ as the variable of integration (the variable that labels the different $dq$). $d\theta$ corresponds to a small change in the angle θ, and is the angle that is subtended by the charge $dq$. That is, the charge $dq$ covers a small arc length, $ds$, of the semi-circle, which is related to $d\theta$ by:

The total charge on the wire is given by $Q$, and the wire has a length $\pi R$ (half the circumference of a circle). Since the charge is distributed uniformly on the wire, the charge per unit length of any piece of wire must be constant. In particular, $dq$ divided by $ds$ must be equal to $Q$ divided by $\pi R$:

where in the last equality we used the relation $ds=Rd\theta$. We now have all of the ingredients to solve the integral:

The total electric field vector at the centre of the circle is thus given by:

Note that if we had not realized that we did not need to solve the integral for the $y$ component, we would still find that it is zero:

In order to determine the electric field at some point from any continuous charge distribution, the procedure is generally the same:

1. Make a good diagram.
2. Choose a charge element $dq$.
3. Draw the electric field element, $d\vec E$, at the point of interest.
4. Write out the electric field element vector, $d\vec E$, in terms of $dq$ and any other relevant variables.
5. Think of symmetry: will any of the component of $d\vec E$ sum to zero over all of the $dq$?
6. Write the total electric field as the sum (integral) of the electric field elements.
7. Identify which variables change as one varies the $dq$ and choose an integration variable to express $dq$ and everything else in terms of that variable and other constants.
8. Do the sum (integral).

15.5The electric dipole¶

Electric dipoles are a specific combination of a positive charge $+Q$ held at a fixed distance, $l$, from an equal and opposite charge, $-Q$, as illustrated in Figure 24.

Dipoles can be represented by their “electric dipole vector” (or “electric dipole moment”), $\vec p$, defined to point in the direction from the negative charge to the positive charge, with magnitude:

Dipoles arise frequently in nature. For example, a water molecule can be modelled as a dipole. In a water molecule, the two hydrogen atoms are not symmetrically arranged around the oxygen atom. The electrons tend to stay closer to the oxygen atom, so the oxygen atom has an excess of 2 electrons, while each proton has a deficit of 1 electron. This results in a separation of charge (polarization), which can be modelled as an electric dipole, as in Figure 25.

When a dipole is immersed in a uniform electric field, as illustrated in Figure 26, the net force on the dipole is zero because the force on the positive charge will always be equal to and in the opposite direction of the force on the negative charge.

Although the net force on the dipole is zero, there is still a net torque about its centre that will cause the dipole to rotate (unless the dipole vector is already parallel to the electric field vector). If the dipole vector makes an angle, θ, with the electric field vector (as in Figure 26), the magnitude of the net torque on the dipole about an axis perpendicular to the page and through the centre of the dipole is given by:

In Figure 26, the torque vector is into the page (the forces will make it rotate clockwise), which is the same direction as the cross product, $\vec p \times \vec E$. Note that the magnitude of the torque is also equal to the magnitude of the cross product. Thus, in general, the torque vector on a dipole, $\vec p$, from an electric field, $\vec E$, is given by:

In particular, note that the torque is zero when the dipole and electric field vectors are parallel. Thus, a dipole will always experience a torque that tends to align it with the electric field vector. The dipole is thus in a stable equilibrium when it is parallel to the electric field.

We can also model the behaviour of the dipole using energy. If a dipole is rotated away from its equilibrium orientation and then released, it will gain (rotational) kinetic energy as it tries to return to equilibrium, and will oscillate about the equilibrium position. When the dipole is held out of equilibrium, we can think of it has having potential energy. To determine the functional form of that potential energy function, we consider the work done in rotating the dipole from an angle $\theta_1$ to an angle $\theta_2$ (where the angle is between the dipole and the electric field vectors):

where the negative sign in the torque is to indicate that the torque is in the opposite direction from increasing θ (in Figure 26, the torque is clockwise whereas the angle θ increases counter-clockwise). The net work done in going from position $\theta_1$ to $\theta_2$ is the negative of the change in potential energy in going from $\theta_1$ to $\theta_2$. Thus, we define the potential energy of an electric dipole, $\vec p$, in an electric field, $\vec E$, as:

which has a negative sign, and we also recognize that this is equivalent to the scalar product between $\vec p$ and $\vec E$. Note that the negative sign makes sense because systems experience a force/torque that will decrease their potential energy. When the angle is zero, $\cos\theta=1$, is maximal. Since we need the position with $\theta=0$ to have the lowest potential energy, the minus sign guarantees that all values of θ other than zero will give a potential energy that is higher (greater than) $(-1) pE$. Remember that only changes in potential energy are relevant, so the minus sign should not bother you, although you should think about whether it makes sense.

15.6Summary¶

Objects can acquire a net charge if they acquire a net excess or deficit of electrons. Charges are never created, they are only transferred from one object to another. One can charge an object by friction, conduction, or induction. Materials can be classified broadly as conductors, where electrons can move freely in a material, or insulators, in which electrons remain tightly bound to the atoms in the material. If a conducting object acquires a net charge, those charges will migrate to the surface of the conductor.

Coulomb was the first to quantitatively describe the electric force exerted on a point charge, $Q_1$, by a second point charge, $Q_2$, located a distance, $r$, away:

where $\hat r_{21}$ is the unit vector from $Q_2$ to $Q_1$. One can write the force using either Coulomb’s constant, $k$, or the permittivity of free space, $\epsilon_0$. Coulomb’s force is attractive if the product $Q_1Q_2$ is negative, and repulsive if the product is positive. Thus, charges of the same sign exert a repulsive force on each other, whereas opposite charges exert an attractive force on each other.

Mathematically, Coulomb’s Law is identical to the gravitational force in Newton’s Universal Theory of Gravity, which implies that it is conservative. The electric field vector at some position in space is defined to be the electric force per unit charge at that position in space. That is, at some position in space where the electric field vector is $\vec E$, a charge, $q$, will experience an electric force:

much like a mass, $m$, will experience a gravitational force, $m\vec g$, in a position in space where the gravitational field is $\vec g$. A positive charge will experience a force in the same direction as the electric field, whereas a negative charge will experience a force in the direction opposite of the electric field. The electric field at position, $\vec r$, from a point charge, $Q$, located at the origin, is given by:

One can visualize an electric field by using “field lines”. The field vector at any point in space has a magnitude that is proportional to the number of field lines at that point, and a direction that is tangent to the field lines at that point.

We can model the electric field from a continuous charged object (i.e. not a point charge) by modelling the object as being made up of many point charges. Often, it is easiest to model an $N$-dimensional object as being the sum of objects of dimension $N-1$ and an infinitesimal length in the remaining dimension. For example, we modelled a line of charge as the sum of point charges that have an infinitesimal length, and we modelled a disk of charge as the the sum of rings that have an infinitesimal thickness. In general, the strategy to model the electric field from a continuous distribution of charge is the same:

1. Make a good diagram.
2. Choose a charge element $dq$.
3. Draw the electric field element, $d\vec E$, at the point of interest.
4. Write out the electric field element vector, $d\vec E$, in terms of $dq$ and any other relevant variables.
5. Think of symmetry: will any of the component of $d\vec E$ sum to zero over all of the $dq$?
6. Write the total electric field as the sum (integral) of the electric field elements.
7. Identify which variables change as one varies the $dq$ and choose an integration variable to express $dq$ and everything else in terms of that variable and other constants.
8. Do the sum (integral).

Finally, we introduced the electric dipole, which is an object comprised of two equal and opposite charges, $+Q$ and $-Q$, held at fixed distance, $l$, from each other. One can model an electric dipole using its dipole vector, $\vec p$, defined to point in the direction from $-Q$ to $+Q$, with magnitude:

When a dipole is immersed in a uniform electric field, $\vec E$, it will experience a torque given by:

The torque will act such as to align the vector $\vec p$ with the electric field vector. We can define a potential energy, $U$, to model the energy that is stored in a dipole when it is not aligned with the electric field:

The point of lowest potential energy corresponds to the case when $\vec p$ and $\vec E$ are parallel, whereas the point of highest potential energy is when the two vectors are anti-parallel.

15.7Thinking about the material¶

15.8Sample problems and solutions¶

15.8.1Problems¶

15.8.2Solutions¶

Solution 15.1

We can model the object as the sum of three finite length wires of the length, $L$. In Example 15.5, we determined that the electric field produced at a distance, $R$, by a finite wire that subtends an angle $2\theta_0$ is given by:

$$E = \frac{2k\lambda}{R}\sin\theta_0$$

(73)

The angle $\theta_0$ is easily found to be $\theta_0 = \frac{\pi}{3}$, as shown in Figure 28. The distance, $R$, is similarly found to be given by:

$$R = \frac{L}{2}\tan\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{6}L$$

(74)

Thus, the magnitude of the electric field from one wire is given by:

$$\begin{align*} E&= \frac{2k\lambda}{R}\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}k\lambda}{R}=\frac{\sqrt{3}k\lambda}{\frac{\sqrt{3}}{6}L}=\frac{6k\lambda}{L} \end{align*}$$

(75)

The charge, $Q$, is evenly distributed along the rod of length, $L$, so that we can rewrite the charge density as $\lambda=\frac{Q}{L}$, which gives:

$$\begin{align*} E &= \frac{6k\lambda}{L} = \frac{6kQ}{L^2} \end{align*}$$

(76)

This is the magnitude of the electric field produced by each side of the triangle. The two positive wires will produce electric fields whose horizontal components ($x$-axis in Figure 28) cancel. The net electric field from the two positive, $\vec E^{pos}$, wires will then be in the negative $y$ direction:

$$E_{y}^{pos}=-2 \left( \frac{6kQ}{L^2} \right) \cos\left(\frac{\pi}{3} \right)=-\frac{6kQ}{L^2}$$

(77)

The negative wire will also produce a field, $\vec E^{neg}$, in the negative $y$ direction (the only component of the field is in the $y$ direction):

$$E_{y}^{neg}=- \frac{6kQ}{L^2}$$

(78)

Thus, the total field at the centre of the triangle, when (vectorially) adding the fields from the two positive wires and the negative wire is given by:

$$\vec E = \left(-\frac{6kQ}{L^2} - \frac{6kQ}{L^2} \right)\hat y=- \frac{12kQ}{L^2}\hat y$$

(79)